${\mathrm{F}}_2 =\frac{-\frac{(110,000 \mathrm{N})(1,500 \mathrm{mm})}{(800 {\mathrm{mm}}^2)(200,000 \mathrm{N}/{\mathrm{mm}}^2)}}{\left(\frac{1,500 \mathrm{mm}}{(800 {\mathrm{mm}}^2)(200,000 \mathrm{N}/{\mathrm{mm}}^2)}+\frac{1,000 \mathrm{mm}}{(1,600 {\mathrm{mm}}^2)(200,000 \mathrm{N}/{\mathrm{mm}}^2)}\right)}$

(Before calculating this value, make a quick check of the units to ensure that they are consistent.)

${\mathrm{F}}_2 =-82,500 \mathrm{N}$ (compression)

Next, solve for ${\mathrm{F}}_1$ from our previous equilibrium equation:

${\mathrm{F}}_1 ={\mathrm{F}}_2+\mathrm{P}=+27,500 \mathrm{N}$ (tension)